Problem: The following function gives the population of a town (in thousands of people), $t$ years after it was founded: $P(t)=20\left(1+e^{10-t}\right)^{-1}$ What is the instantaneous rate of change of the population, $6$ years after the town was founded? Choose 1 answer: Choose 1 answer: (Choice A) A $0.06$ people per year (Choice B) B $0.06$ thousand people per year (Choice C) C $0.35$ people per year (Choice D) D $0.35$ thousand people per year
Explanation: Understanding the problem The instantaneous rate of change of $P(t)$ is given by its derivative, $P'(t)$. Therefore, the instantaneous rate of change of the population after $6$ years is $P'(6)$. Let's find $P'(t)$ and evaluate it at $t=6$. Finding $P'(t)$ $P'(t)=20e^{10-t}\left(1+e^{10-t}\right)^{-2}$ Finding $P'(6)$ $\begin{aligned} P'({6})&=20e^{10-({6})}\left(1+e^{10-({6})}\right)^{-2} \\\\ &=\dfrac{20e^4}{(1+e^4)^2} \\\\ &\approx 0.35 \end{aligned}$ Interpreting units $P(t)$ is the population in ${\text{thousands of people}}$ after $t$ ${\text{years}}$. Therefore, we measure its rate of change in ${\text{thousands of people}}$ per ${\text{year}}$. In conclusion, the instantaneous rate of change of the population, $6$ years after the town was founded, is $0.35$ thousand people per year. The rate of change is positive because the size of the population is increasing.